In a proof I am reading, the author considers the situation where $G$ is a reductive algebraic group (variety) over the complex numbers $\mathbb C$ and $N\trianglelefteq G$ is a closed, normal subgroup of $G$ which is also finite. The author concludes that "because $G$ is reductive", $N$ must be contained in the center of $G$. I can't seem to make that same deduction, though. Is this some well-known result? Can you give a proof?
Btw, if $G$ is connected then this does not seem to have anything to do with reductivity at all. Let $u\in N$, then $G.u\subseteq N$ is finite. Since $0=\dim(G.u)=\dim(G)-\dim(G_u)$, the stabilizer of $u$ is a closed subgroup of $G$ which has maximal dimension, therefore $G=G_u$. However, the author explicitly states that the reductivity of $G$ is the reason for his deduction, and I am curious what that would be.
I recall writing about this a while ago. This holds for connected algebraic groups over an algebraically closed field $k$, not necessarily reductive: Let $H$ be the normal subgroup. Let $y\in H$ and let $G\to G$ be given by $x\mapsto xyx^{-1}$. Since $H$ is normal, this sends $G$ into $H$. The image of a connected group is connected, so the image is a single point. In particular, taking $x = 1_G$, we see that the image of $G$ is actually $y$. Hence $y$ commutes with every element of $G$.
If you take the definition of $G$ being reductive as just trivial unipotent radical, this is surely false. For instance, $N = G$, and $G$ a nonabelian finite group that is an algebraic group.